INTERPRETING THE INDEPENDENT-SAMPLES t TEST
The first table, (Group Statistics) shows descriptive statistics for the two groups (low-stress
and high-stress) separately. Note that the means for the two groups look somewhat different.
This might be due to chance, so we will want to test this with the t test in the next table.
The second table, (Independent Samples Test) provides two statistical tests. In the left two
columns of numbers, is the Levene’s Test for Equality of Variances for the assumption
that the variances of the two groups are equal (i.e., assumption of homogeneity of variance).
NOTE, this is not the t test; it only assesses an assumption! If this F test is not significant (as in
the case of this example), the assumption is not violated (that is, the assumption is met), and one
uses the Equal variances assumed line for the t test and related statistics. However, if
Levene’s F is statistically significant (Sig < 0.05), then variances are significantly different and
the assumption of equal variances is violated (not met). In that case, the Equal variances not
assumed line would be used – for which SPSS adjusts the t, df, and Sig. as appropriate.
Also in the second table… we obtain the needed information to test the equality of the means.
Recall that there are three methods in which we can make this determination.
METHOD ONE (most commonly used): comparing the Sig. (probability) value (p = .022 for
our example) to the a priori alpha level ( = 0.05 for our example). If p <  – we reject the null
hypothesis of no difference. If p >  – we retain the null hypothesis of no difference. For our
example, p < , therefore we reject the null hypothesis and conclude that the low-stress group (M
= 45.20) talked significantly more than did the high-stress group (M = 22.07).
METHOD TWO: comparing the obtained t statistic value (tobt = 2.430 for our example) to the t
critical value (tcv). Knowing that we are using a two-tailed (non-directional) t test, with an alpha
level of 0.05 ( = 0.05), with df = 28, and looking at the Student’s t Distribution Table – we find
the critical value for this example to be 2.048. If |tobt| > |tcv| – we reject the null hypothesis of no
difference. If |tobt| < |tcv| – we retain the null hypothesis of no difference. For our example, tobt =
2.430 and tcv = 2.048, therefore, tobt > tcv – so we reject the null hypothesis and conclude that
there is a statistically significant difference between the two groups. More specifically, looking
at the group means, we conclude that the low-stress group (M = 45.20) talked significantly more
than did the high-stress group (M = 22.07).
INTERPRETING THE INDEPENDENT-SAMPLES t TEST
METHOD THREE: examining the confidence intervals and determining whether the upper (42.637
for our example) and lower (3.630 for our example) boundaries contain zero (the hypothesized
mean difference). If the confidence intervals do not contain zero – we reject the null hypothesis
of no difference. If the confidence intervals do contain zero – we retain the null hypothesis of no
difference. For our example, the confidence intervals (+3.630, +42.637) do not contain zero,
therefore we reject the null hypothesis and conclude that the low-stress group (M = 45.20) talked
significantly more than did the high-stress group (M = 22.07).
Note that if the upper and lower bounds of the confidence intervals have the same sign
(+ and + or – and –), we know that the difference is statistically significant because this
means that the null finding of zero difference lies outside of the confidence interval.
CALCULATING
AN
EFFECT SIZE: Since we concluded that there was a significant difference
between the average amount of time spent talking between the two groups – we will need to
calculate an effect size to determine the magnitude of this significant effect. Had we not found a
significant difference – no effect size would have to be calculated (as the two groups would have
only differed due to random fluctuation or chance).
To calculate the effect size for this example, we will use the following formula:
d t
Where,
N1  N 2
N1 N 2
t = 2.43
N1 = 15
N2 = 15
Substituting the values into the formula – we find:
d  2.43
15  15
30
 2.43
 2.43 .133333  (2.43)(. 365148)  .887310 = .89
(15)(15)
225
INTERPRETING THE INDEPENDENT-SAMPLES t TEST
Independent-Samples T-Test Example
Group Statistics
TALK
STRESS
1 Low Stress
2 High Stress
N
15
15
Mean
45.20
22.07
Std. Deviation
24.969
27.136
Std. Error
Mean
6.447
7.006
Independent Samples Test
Levene's Test for
Equality of Variances
F
TALK
Equal variances
as sumed
Equal variances
not ass umed
.023
Sig.
.881
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
2.430
28
.022
23.13
9.521
3.630
42.637
2.430
27.808
.022
23.13
9.521
3.624
42.643
Drop-Down Syntax for Independent-Samples T-Test Example
T-TEST
GROUPS=stress(1 2)
/MISSING=ANALYSIS
/VARIABLES=talk
/CRITERIA=CIN(.95) .
Alternative Syntax for Independent-Samples T-Test Example
t-test / groups = stress (1, 2) / variables = talk.